Count Operations to Obtain Zero
Here’s the code that implements the logic for counting the number of operations required to make either num1
or num2
zero. The given constraints and conditions are handled in the code.
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 class Solution:
def countOperations(self, num1: int, num2: int) > int:
# Initialize a counter to keep track of the number of operations
operations_count = 0
# Continue the loop until either num1 or num2 becomes zero
while num1 != 0 and num2 != 0:
# If num1 is greater than or equal to num2, subtract num2 from num1
if num1 >= num2:
# Determine how many times num2 fits into num1
times = num1 // num2
# Subtract num2 from num1 that many times
num1 = num2 * times
# Increment the operations count by that many times
operations_count += times
else: # If num2 is greater than num1, subtract num1 from num2
# Determine how many times num1 fits into num2
times = num2 // num1
# Subtract num1 from num2 that many times
num2 = num1 * times
# Increment the operations count by that many times
operations_count += times
return operations_count

The code loops until either num1
or num2
is zero, and in each iteration, it subtracts the smaller number from the larger number, updating the count accordingly. The use of the division operator //
allows us to perform the subtraction multiple times in a single step if the numbers are multiples of each other, thereby optimizing the code.